The sum of the series binom{20}{0} - binom{20 | vedclass

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(D) Let $S = \binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \dots + \binom{20}{10}$.We know that $\binom{n}{r} = \binom{n}{n-r}$.Thus,$\binom{20}{0}

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$\frac{1}{2} \binom{20}{10}$

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(D) Let $S = \binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \dots + \binom{20}{10}$.We know that $\binom{n}{r} = \binom{n}{n-r}$.Thus,$\binom{20}{0} = \binom{20}{20}$,$\binom{20}{1} = \binom{20}{19}$,...,$\binom{20}{9} = \binom{20}{11}$.Consider the expansion $(1-1)^{20} = \sum_{r=0}^{20} (-1)^r \binom{20}{r} = 0$.This gives $\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9} + \binom{20}{10} - \binom{20}{11} + \dots + \binom{20}{20} = 0$.Using symmetry,$\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9} + \binom{20}{10} - \binom{20}{9} + \dots + \binom{20}{0} = 0$.$2[\binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9}] + \binom{20}{10} = 0$.Let $X = \binom{20}{0} - \binom{20}{1} + \dots - \binom{20}{9}$. Then $2X + \binom{20}{10} = 0$,so $X = -\frac{1}{2} \binom{20}{10}$.The required sum is $S = X + \binom{20}{10} = -\frac{1}{2} \binom{20}{10} + \binom{20}{10} = \frac{1}{2} \binom{20}{10}$.

The sum of the series $\binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \binom{20}{3} + \dots + \binom{20}{10}$ is:

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